Answer

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Hint: To find the number of three digit numbers that have 3 in the unit place, consider all possible combinations of numbers allowed to be placed at hundreds and tens place, keeping in mind whether repetition is allowed or not. Multiply the value of the number of digits allowed at both tens and hundreds place to count the total number of possible combinations.

Complete step-by-step answer:

We have to count the possible three digit numbers which have 3 in unit place, in both the cases when repetition of digits is allowed and repetition is not allowed.

We will count all possible combinations in each case.

(i) We will firstly count the total combinations when repetition of digits is allowed.

Thus, we can have all the digits from \[1-9\] at the hundreds place. At tens place, we can have any of the digits from \[0-9\].

So, 9 different digits can be placed at hundreds places and 10 different digits can be placed at hundreds places.

Thus, the total number of possible combinations of digits to form a three digit number \[=9\times 10=90\].

Hence, 90 different combinations of digits are allowed to form three digit numbers having 3 at units place when repetition of digits is allowed.

(ii) We will now calculate the number of possible combinations to form three digit numbers when repetition of digits is not allowed. Thus, we can place each digit only once at any position in the number.

In this case, we will consider two possible sub cases – when 0 is placed at tens place and 0 is not placed at tens place.

If 0 is placed at tens place, we only have to count the arrangement of digits at hundreds place. At hundreds places, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\].

So, 8 digits are allowed to be placed at hundreds places.

Thus, 8 three digit numbers are possible which have 3 at units place and 0 at tens place.

We will now count the possible number of three digit numbers which don’t have 0 at tens place. At hundreds places, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\]. At tens place, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\].

However, we need to remove any one of the digits from tens place as that digit will already be placed at hundreds place.

So, the number of digits allowed at hundreds places is 8 and the number of digits allowed at tens places is 7.

Total number of three digits numbers which have 3 at unit place and don’t have 0 at tens place \[=8\times 7=56\].

Thus, the total number of possible three digit numbers which have 3 at units place without repetition \[=56+8=64\].

Hence, 64 different combinations of digits are allowed to form three digit numbers having 3 at units place when repetition of digits is not allowed.

Note: It’s necessary to keep in mind that we can’t place 0 at hundreds places. Otherwise, we will get two digit numbers. Also, one shouldn’t count the digit 3 when repetition is not allowed as we can have any digit only once. However, when repetition is allowed, we can count each digit multiple times. We can also solve this question by using the combination formula \[{}^{n}{{C}_{r}}\] to choose r digits from a set of n digits.

Complete step-by-step answer:

We have to count the possible three digit numbers which have 3 in unit place, in both the cases when repetition of digits is allowed and repetition is not allowed.

We will count all possible combinations in each case.

(i) We will firstly count the total combinations when repetition of digits is allowed.

Thus, we can have all the digits from \[1-9\] at the hundreds place. At tens place, we can have any of the digits from \[0-9\].

So, 9 different digits can be placed at hundreds places and 10 different digits can be placed at hundreds places.

Thus, the total number of possible combinations of digits to form a three digit number \[=9\times 10=90\].

Hence, 90 different combinations of digits are allowed to form three digit numbers having 3 at units place when repetition of digits is allowed.

(ii) We will now calculate the number of possible combinations to form three digit numbers when repetition of digits is not allowed. Thus, we can place each digit only once at any position in the number.

In this case, we will consider two possible sub cases – when 0 is placed at tens place and 0 is not placed at tens place.

If 0 is placed at tens place, we only have to count the arrangement of digits at hundreds place. At hundreds places, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\].

So, 8 digits are allowed to be placed at hundreds places.

Thus, 8 three digit numbers are possible which have 3 at units place and 0 at tens place.

We will now count the possible number of three digit numbers which don’t have 0 at tens place. At hundreds places, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\]. At tens place, we can have any one of the digits from the set \[\left\{ 1,2,4,5,6,7,8,9 \right\}\].

However, we need to remove any one of the digits from tens place as that digit will already be placed at hundreds place.

So, the number of digits allowed at hundreds places is 8 and the number of digits allowed at tens places is 7.

Total number of three digits numbers which have 3 at unit place and don’t have 0 at tens place \[=8\times 7=56\].

Thus, the total number of possible three digit numbers which have 3 at units place without repetition \[=56+8=64\].

Hence, 64 different combinations of digits are allowed to form three digit numbers having 3 at units place when repetition of digits is not allowed.

Note: It’s necessary to keep in mind that we can’t place 0 at hundreds places. Otherwise, we will get two digit numbers. Also, one shouldn’t count the digit 3 when repetition is not allowed as we can have any digit only once. However, when repetition is allowed, we can count each digit multiple times. We can also solve this question by using the combination formula \[{}^{n}{{C}_{r}}\] to choose r digits from a set of n digits.

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